POJ 1222 EXTENDED LIGHTS OUT（高斯消元）题解

题意：5*6的格子，你翻一个地方，那么这个地方和上下左右的格子都会翻面，要求把所有为1的格子翻成0，输出一个5*6的矩阵，把要翻的赋值1，不翻的0，每个格子只翻1次

思路：poj 1222 高斯消元详解

代码：

#include<queue>
#include<cstring>
#include<set>
#include<map>
#include<stack>
#include<cmath>
#include<vector>
#include<cstdio>
#include<iostream>
#include<algorithm>
#define eps 1e-9
typedef long long ll;
const int maxn = 1e4 + 10;
const int seed = 131;
const ll MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
using namespace std;
int a[35][35], x[35], Case = 1;
int equ, var;
int free_num,free_x[35];
int Gauss(){
    int max_r, col, k;
    free_num = 0;
    equ = var = 30;
    for(k = 0, col = 0; k < equ && col < var; k++, col++){
        max_r = k;
        for(int i = k + 1; i < equ; i++){
            if(abs(a[i][col]) > abs(a[max_r][col]))
                max_r = i;
        }
        if(a[max_r][col] == 0){
            k--;
            free_x[free_num++] = col;
            continue;
        }
        if(max_r != k){
            for(int j = col; j < var + 1; j++){
                swap(a[k][j], a[max_r][j]);
            }
        }
        for(int i = k + 1; i < equ; i++){
            if(a[i][col] != 0){
                for(int j = col; j < var + 1; j++){
                    a[i][j] ^= a[k][j];
                }
            }
        }
    }
    for(int i = k; i < equ; i++){
        if(a[i][col] != 0)
            return -1;
    }
    if(k < var) return var - k;
    for(int i = var - 1; i >= 0; i--){
        x[i] = a[i][var];
        for(int j = i + 1; j < var; j++){
            x[i] ^= (a[i][j] & x[j]);
        }
    }
    return 0;
}
int pos(int i, int j){
    return i * 6 + j;
}
void solve(){
    int u;
    memset(a, 0, sizeof(a));
    memset(x, 0, sizeof(x));
    for(int i = 0; i < 5; i++){
        for(int j = 0; j < 6; j++){
            if(i > 0) a[pos(i - 1, j)][pos(i, j)] = 1;
            if(i < 4) a[pos(i + 1, j)][pos(i, j)] = 1;
            if(j > 0) a[pos(i, j - 1)][pos(i, j)] = 1;
            if(j < 5) a[pos(i, j + 1)][pos(i, j)] = 1;
            a[pos(i, j)][pos(i, j)] = 1;
            int u;
            scanf("%d", &u);
            a[pos(i, j)][30] = u;
        }
    }
    Gauss();
    printf("PUZZLE #%d\n", Case++);
    for(int i = 0; i < 5; i++){
        for(int j = 0; j < 6; j++){
            if(j != 0) printf(" ");
            printf("%d", x[pos(i, j)]);
        }
        printf("\n");
    }
}
int main(){
    int T;
    scanf("%d", &T);
    while(T--){
        solve();
    }
    return 0;
}